Official solution
When 2x \[ \ge \] 1
5 + 2x --1 = 2x (2x -- 2)
Let 2x = t
\[ \Rightarrow \]5 + t -- 1 = t (t -- 2)
\[ \Rightarrow \] t = 4, -- 1(rejected)
\[ \Rightarrow \] 2x = 4
\[ \Rightarrow \] x = 2
Now when 2x < 1
5 + 1 -- 2x = 2x (2x -- 2)
Let 2x = t
\[ \Rightarrow \] 5 + 1 -- t = t (t -- 2)
\[ \Rightarrow \] 0 = t2 -- t -- 6
\[ \Rightarrow \] 0 = (t -- 3) (t -- 2)
\[ \Rightarrow \] t = 3, -- 2
2x = 3, 2x = -- 2 (rejected)
\[ \therefore \] Only one real roots.